Random Eigenvalue Problems in Structural Dynamics S ONDIPON A - - PowerPoint PPT Presentation

Random Eigenvalue Problems in Structural Dynamics

SONDIPON ADHIKARI

Department of Aerospace Engineering University of Bristol Bristol, U.K. Email: S.Adhikari@bristol.ac.uk URL: http://www.aer.bris.ac.uk/contact/academic/adhikari/home.html

Random Eigenvalue Problems – p.1/21

Outline of the talk

Random eigenvalue problem Perturbation Methods Mean-centered perturbation method α-centered perturbation method Asymptotic analysis Numerical Example Conclusions & Open Problems.

Random Eigenvalue Problems – p.2/21

Random eigenvalue problem

The random eigenvalue problem of undamped or proportionally damped linear systems: K(x)φj = λjM(x)φj. (1) λj eigenvalues; φj eigenvectors; M(x) ∈ RN×N mass matrix and K(x) ∈ RN×N stiffness matrix. x ∈ Rm is random parameter vector with pdf p(x) = (2π)−m/2e−xTx/2. (2)

Random Eigenvalue Problems – p.3/21

The fundamental aim

To obtain the joint probability density function of the eigenvalues and the eigenvectors. If the matrix M−1K is GUE (Gaussian unitary ensemble) or GOE (Gaussian orthogonal ensemble) an exact closed-form expression can be obtained for the joint pdf of the eigenvalues. In general the system matrices for real structures are not GUE or GOE

Random Eigenvalue Problems – p.4/21

Mean-centered perturbation method

Assume that M(0) = M0 and K(0) = K0 are ‘deterministic parts’ (in general different from the mean matrices). The deterministic eigenvalue problem K0φj0 = λj0M0φj0. The eigenvalues λj(x) : Rm → R are non-linear functions of

• x. Here λj(x) is replaced by its Taylor series about the

point x = 0 λj(x) ≈ λj(0) + dT

λj(0)x + 1

2xTDλj(0)x. (3) dλj(0) ∈ Rm and Dλj(0) ∈ Rm×m are respectively the gra- dient vector and the Hessian matrix of λj(x) evaluated at x = 0.

Random Eigenvalue Problems – p.5/21

α-centered perturbation method

We are looking for a point x = α in the x-space such that the Taylor series expansion of λj(x) about this point λj(x) ≈ λj(α) + dT

λj(α) (x − α) + 1

2 (x − α)T Dλj(α) (x − α) (4) is optimal in some sense. The optimal point α is selected such that the mean or the first moment of each eigenvalue is calculated most accurately.

Random Eigenvalue Problems – p.6/21

α-centered perturbation method

The mean of λj(x) can be obtained as ¯ λj =

• R

m λj(x)p(x) dx = (2π)−m/2

• R

m e−h(x) dx

(5) where h(x) = xTx/2 − ln λj(x). (6) Expand the function h(x) in a Taylor series about a point where h(x) attends its global minimum. By doing so the error in evaluating the integral (5) would be minimized. Therefore, the optimal point can be obtained as ∂h(x) ∂xk = 0

• r

xk = 1 λj(x) ∂λj(x) ∂xk , ∀k. (7)

Random Eigenvalue Problems – p.7/21

α-centered perturbation method

Combining for all k we have dλj(α) = λj(α)α. Rearranging α = dλj(α)/λj(α). (8) This equation immediately gives a recipe for an iterative algorithm to obtain α. Substituting dλj(α) in Eq. (4) λj(x) ≈ λj(α)

• 1 − |α|2

+ 1 2αTDλj(α)α + αT λj(α)I − Dλj(α)

• x + 1

2xTDλj(α)x. (9)

Random Eigenvalue Problems – p.8/21

Eigenvalue statistics using theory of quadratic forms

Both approximations yield a quadratic form in Gaussian random variable λj(x) ≈ cj + aT

j x + 1 2xTAjx.

The moment generating function: Mλj(s) = E

• esλj(x)

≈ escj+ s2

2 aT j [I−sAj] −1aj

• I − sAj

(10) Cumulants: κr =

• cj + 1

2Trace (Aj)

if r = 1,

r! 2 aT j Ar−2 j

aj + (r−1)!

2

Trace

• Ar

j

• if

r ≥ 2. (11)

Random Eigenvalue Problems – p.9/21

Asymptotic analysis

We want to evaluate an integral of the following form: J =

• R

m f(x)p(x) dx = (2π)−m/2

• R

m e

• h(x) dx

(12) where

• h(x) = ln f(x) − xTx/2.

(13) Assume f(x) : Rm → R is smooth and at least twice differentiable and h(x) reaches its global maximum at an unique point θ ∈ Rm. Therefore, at x = θ ∂ h(x) ∂xk = 0 or xk = ∂ ∂xk ln f(x), ∀k, or θ = ∂ ∂x ln f(θ). (14)

Random Eigenvalue Problems – p.10/21

Asymptotic analysis

Further assume that h(θ) is so large that

• 1
• h(θ)

Dj( h(θ))

• → 0

for j > 2 (15) where Dj( h(θ)) is jth order derivative of h(x) evaluated at x = θ. Under such assumptions, using second-order Taylor series of h(x) the integral (12) can be evaluated as J ≈ e

• h(θ)
• H(θ)

= f(θ)e

− θ

Tθ/2

• H(θ)−1/2.

(16)

Random Eigenvalue Problems – p.11/21

Asymptotic analysis

An arbitrary rth order moment of the eigenvalues µ′

r =

• R

m λr

j(x)p(x) dx,

r = 1, 2, 3 · · · (17) Comparing this with Eq. (12) it is clear that f(x) = λr

j(x)

and

• h(x) = r ln λj(x) − xTx/2.

(18) The optimal point θ can be obtained from (14) as θ = r dλj(θ)/λj(θ). (19)

Random Eigenvalue Problems – p.12/21

Asymptotic analysis

Using the asymptotic approximation, the rth moment: µ′

r = λr j(θ)e− |θ|2

2

• I + 1

rθθT − r λj(θ)Dλj(θ)

• −1/2

. (20) The mean of the eigenvalues (by substituting r = 1): ¯ λj = λj(θ)e− |θ|2

2

• I + θθT − Dλj(θ)/λj(θ)
• −1/2 .

(21) Central moments: E

• (λj − ¯

λj)r = r

k=0

r

k

• (−1)r−kµ′

k¯

λr−k

j

.

Random Eigenvalue Problems – p.13/21

Numerical example

Undamped two degree-of-system system: m1 = 1 Kg, m2 = 1.5 Kg, ¯ k1 = 1000 N/m, ¯ k2 = 1100 N/m and k3 = 100 N/m.

m

1

m

2 1 2

k

1

k

2

k

3

Only the stiffness parameters k1 and k2 are uncertain: ki = ¯ ki(1 + ǫixi), i = 1, 2. x = {x1, x2}T ∈ R2 and the ‘strength parameters’ ǫ1 = ǫ2 = 0.25.

Random Eigenvalue Problems – p.14/21

Numerical example

Following six methods are compared

• 1. Mean-centered first-order perturbation
• 2. Mean-centered second-order perturbation
• 3. α-centered first-order perturbation
• 4. α-centered second-order perturbation
• 5. Asymptotic method
• 6. Monte Carlo Simulation (10K samples) - can be

considered as benchmark. The percentage error: Errorith method = {µ′

k}ith method − {µ′ k}MCS

{µ′

k}MCS

× 100 .

Random Eigenvalue Problems – p.15/21

Numerical example

1 2 3 4 2 4 6 8 10 12 14 16 18 20 k−th order moment: E [λk

1]

Percentage error wrt MCS Mean−centered 1st−order Mean−centered 2nd−order α−centered 1st−order α−centered 2nd−order Asymptotic Method

Percentage error for the first four raw moments of the first eigenvalue

Random Eigenvalue Problems – p.16/21

Numerical example

1 2 3 4 −20 −18 −16 −14 −12 −10 −8 −6 −4 −2 k−th order moment: E [λk

2]

Percentage error wrt MCS Mean−centered 1st−order Mean−centered 2nd−order α−centered 1st−order α−centered 2nd−order Asymptotic Method

Percentage error for the first four raw moments of the second eigenvalue

Random Eigenvalue Problems – p.17/21

Numerical example

500 1000 1500 0.5 1 1.5 2 2.5 3 x 10−3 u pλ

1

(u) Mean−centered 1st−order Mean−centered 2nd−order α−centered 1st−order α−centered 2nd−order Asymptotic Method

Probability density function of the first eigenvalue

Random Eigenvalue Problems – p.18/21

Numerical example

400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x 10−3 u pλ

2

(u) Mean−centered 1st−order Mean−centered 2nd−order α−centered 1st−order α−centered 2nd−order Asymptotic Method

Probability density function of the second eigenvalue

Random Eigenvalue Problems – p.19/21

Conclusions

Two methods, namely (a) optimal point expansion method, and (b) asymptotic moment method, are proposed. The optimal point is obtained so that the mean of the eigenvalues are estimated most accurately. The asymptotic method assumes that the eigenvalues are large compared to their 3rd order or higher derivatives. Pdf of the eigenvalues are obtained in terms of central and non-central χ2 densities.

Random Eigenvalue Problems – p.20/21

Open problems

Joint statistics (moments/pdf/cumulants) of the eigenvalues with non-Gaussian system parameters. Statistics of the difference and ratio of the eigenvalues. Statistics of a single eigenvector (for GUE/GOE and general matrices). Joint statistics of the eigenvectors. Joint statistics of the eigenvalues and eigenvectors.

Random Eigenvalue Problems – p.21/21